EXPLANATION OF LITHIUM IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) April 24, 2015 Lithium is a chemical element with symbol Li and atomic number 3. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this image of Lithium including the following ground state electron configuration: 1s22s1. According to the “Ionization energies of the elements-WIKIPEDIA” we observe that E1 = 5.39 eV, E2 = 75.64 eV, and E3 = 122.4 eV. The first ionization energy (E1 = 5.39 eV) is due to the outer electron (2s1) with n = 2. That is E1 = 0 - E( 2s1) = 0 - ( -5.39) = 5.39 eV Whereas for calculating the second ionization energy (E2 = 75.64 eV) in my paper of 2008 I showed that the second E2 = 75. 64 eV is given by E2 = E(1s1) - E(1s2) Here E(1s1) is the binding energy of the one electron of Li2+ with n = 1 based on the Bohr model, while E(1s2) is the binding energy of the two spinning electrons of opposite spin of the Li+ with n = 1 based on my formula of 2008: E2 = (-13.6) Z2 - [ (-27.2)Z2 +(16.95)Z - 4.1] Since Z = 3 one gets E2 = (-13.6)32 + (27.2)32 - (16.95)3 + 4.1 = - 122.4 + 244.8 - 50.84 + 4.1 = 75.64 eV Of course, the third ionization energy E3 = 122.4 eV is due to the one remaining electron of 1s1 with n =1 given by applying the simple Bohr model for Z = 3 as E3 = 0 - (-13.6)Z2 = 0 - (-13.6)32 = 122.4 eV ' ' EXPLANATION OF THE FIRST IONIZATION ENERGY E1 = 5.39 eV ' In my paper of 2008, I showed that the two electrons of opposite spin of the 1s2 spherical shell with n = 1 and the third spinning electron of 2s1 spherical shell with n = 2 interact electrically with a Coulomb repulsion. So the outer third electron of 2s1 doesn't feel the full pull of the nucleus with Z = 3, because it is screened by the two electrons of the shell 1s2 . For a perfect screening, due to spherical symmetries, since the outer electron with n = 2 looks inward at just one net positive charge, ( effective charge Zeff = ζ =1) it could be expected to have energy level similar to that of hydrogen given by E = (-13.6)ζ2/n2 = (-13.6)12/22 = -3.4 eV This is true for high angular momentum states, but according to the quantum mechanics the s and p states fall well below the corresponding hydrogen energy levels. The lithium 2s level is significantly lower than the 2p because of greater penetration past the shielding of the 1s2 two electrons of spherical symmetry. Both levels of s and p penetrate enough to be significantly lower than the n = 2 hydrogen energy which they would have if the shielding were perfect. Also in my paper of 2008, I showed that the penetrating corresponds to the greater repulsion between the 1s2 and 2s1 which leads to the great deformation of spherical symmetries of both 1s2 shell with n = 1 and 2s1 shell with n = 2 . It means that the effective ζ >1 . Thus from the ionization energy E1 = 5.39 eV using the Bohr model for the one outer electron 2s1 with n = 2 one can calculate the value of the effective ζ as E1 = 0 - (-13.6) ζ2/22 = 5.39 eV That is ζ2 = 1.585294 and ζ = 1,26 > 1 . ' ' ' EXPLANATION OF THE SECOND IONIZATION ENERGY E2 = 75.64 eV In my paper of 2008, I showed that the second ionization energy E2 is given by E2 = E( 1s1) - E(1s2 ) = (-13.6) Z2 - [ (-27.2)Z2 +(16.95)Z - 4.1] Since Z = 3 one gets E2 = (-13.6)32 + (27.2)32 - (16.95)3 + 4.1 = - 122.4 + 244.8 - 50.84 + 4.1 = 75.64 eV Here the E(1s1) with n = 1 is given just by applying the Bohr formula, while the E(1s2) of two spinning electrons of opposite spin with n =1 is given by applying my formula of 2008. However In the absence of a detailed knowledge about the electromagnetic force between the two spinning electrons of opposite spin physicist today using wrong theories cannot explain the ionization energy E2 = 75.64 eV. For example under wrong theories based on qualitative approaches many physicists believe incorrectly that the second electron of the 1s2 shell is less tightly bound because it could be interpreted as a shielding effect; the other electron partly shields the second electron from the full charge of the nucleus. Another wrong way to view the energy is to say that the repulsion of the electrons contributes a positive potential energy which partially offsets the negative potential energy contributed by the attractive electric force of the nuclear charge”. In atomic physics a two-electron atom is a quantum mechanical system consisting of one nucleus with a charge Ze and just two electrons. This is the first case of many-electron systems. The first few two-electron atoms are: Z =1 : H- hydrogen anion. Z = 2 : He helium atom. Z = 3 : Li+ lithium atom anion. Z = 4 : Be2+ beryllium ion. Z = 5 : B3+ boron. Prior to the development of quantum mechanics, an atom with many electrons was portrayed like the solar system, with the electrons representing the planets circulating about the nuclear “sun”. In the solar system, the gravitational interaction between planets is quite small compared with that between any planet and the very massive sun; interplanetary interactions can, therefore, be treated as small perturbations. However, In the helium atom with two electrons, the interaction energy between the two spinning electrons and between an electron and the nucleus are almost of the same magnitude, and a perturbation approach is inapplicable. In 1925 the two young Dutch physicists Uhlenbeck and Goudsmit discovered the electron spin according to which the peripheral velocity of a spinning electron is greater than the speed of light. Since this discovery invalidates Einstein’s relativity it met much opposition by physicists including Pauli. Under the influence of Einstein’s invalid relativity physicists believed that in nature cannot exist velocities faster than the speed of light.(See my FASTER THAN LIGHT). So great physicists like Pauli, Heisenberg, and Dirac abandoned the natural laws of electromagnetism in favor of wrong theories including qualitative approaches under an idea of symmetry properties between the two electrons of opposite spin which lead to many complications. Thus, in the “Helium atom-Wikipedia” one reads: “Unlike for hydrogen a closed form solution to the Schrodinger equation for the helium atom has not been found. However various approximations such as the Hartree-Fock method ,can be used to estimate the ground state energy and wave function of atoms”. It is of interest to note that in 1993 in Olympia of Greece I presented at the international conference “Frontiers of fundamental physics” my paper “Impact of Maxwell’s equation of displacement current on electromagnetic laws and comparison of the Maxwellian waves with our model of dipolic particles ". The conference was organized by the natural philosophers M. Barone and F. Selleri who awarded me an award including a disc of the atomic philosopher Democritus because In that paper I showed that LAWS AND EXPERIMENTS INVALIDATE FIELDS AND RELATIVITY . At the same period I tried to find not only the nuclear force and structure but also the coupling of two electrons under the application of the abandoned electromagnetic laws. For example in the well known photoelectric effect the absorption of light contributes not only to the increase of the electron energy but also to the increase of the electron mass, because the particles of light have mass m = hν/c2 .( See my DISCOVERY OF PHOTON MASS ). However the electron spin which gives a peripheral velocity greater than the speed of light cannot be affected by the photon absorption. Thus after 10 years I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism" (2003), in which I showed not only my DISCOVERY OF NUCLEAR FORCE AND STRUCTURE but also that the peripheral velocity (u >> c) of two spinning electrons with opposite spin gives an attractive magnetic force (Fm) stronger than the electric repulsion (Fe) when the two electrons of mass m and charge (-e) are at a very short separation ( r < 578.8 /1015 m) . Because of the antiparallel spin along the radial direction the interaction of the electron charges gives an electromagnetic force Fem = Fe - Fm . Therefore in my research the integration for calculating the mutual Fem led to the following relation: Fem = Fe - Fm = Ke2/r2 - (Ke2/r4)(9h2/16π2m2c2) Of course for Fe = Fm one gets the equilibrium separation ro = 3h/4πmc = 578.8/1015 m. That is, for r < 578.8/1015 m the two electrons of opposite spin exert an attractive electromagnetic force, because the attractive Fm is stronger than the repulsive Fe . Here Fm is a spin-dependent force of short range. As a consequence this situation provides the physical basis for understanding the pairing of two electrons described qualitatively by the Pauli principle, which cannot be applied in the simplest case of the deuteron in nuclear physics, because the binding energy between the two spinning nucleons occurs when the spin is not opposite (S=0) but parallel (S=1). According to the experiments in the case of two electrons with antiparallel spin the presence of a very strong external magnetic field gives parallel spin (S=1) with electric and magnetic repulsions given by Fem = Fe + Fm So, according to the well-established laws of electromagnetism after a detailed analysis of paired electrons in two-electron atoms I concluded that at r < 578.8/1015 m a motional EMF produces vibrations of paired electrons. Unfortunately today many physicists in the absence of a detailed knowledge believe that the two electrons of two-electron atoms under the Coulomb repulsion between the electrons move not together as one particle but as separated particles possessing the two opposite points of the diameter of the orbit around the nucleus. In fact, the two electrons of opposite spin behave like one particle circulating about the nucleus under the rules of quantum mechanics forming two-electron orbitals in helium, beryllium etc. In my paper of 2008, I showed that the positive vibration energy (Ev) described in eV depends on the Ze charge of nucleus as Ev = (16.95)Z - 4.1 Of course in the absence of such a vibration energy Ev it is well-known that the ground state energy E described in eV for two orbiting electrons could be given by the Bohr model as E = (-27.2) Z2. So the combination of the energies of the Bohr model and the vibration energies due to the opposite spin of two electrons led to my discovery of the ground state energy of two-electron atoms given by E = (-27.2) Z2 + (16.95 )Z - 4.1 For example the laboratory measurement of the ionization energy of H- yields an energy of the ground state E = - 14.35 eV. In this case since Z = 1 we get E -27.2 + 16.95 - 4.1 = -14.35 eV. In the same way writing for the helium Z = 2 we get E = - 108.8 + 32.9 - 4.1 = -79.0 eV which is equal to the laboratory measurement. In the same way we can calculate the ground state energies for the Z = 3 : Li+ ion , Z = 4 : Be2+ beryllium ion, and Z = 5 : B3+ boron. The discovery of this simple formula based on the well-established laws of electromagnetism was the first fundamental equation for understanding the energies of many-electron atoms, while various theories based on qualitative symmetry properties lead to complications. CONCLUSIONS To explain the first ionization energy E1 = 5.39 eV one must find the effective Zeff = ζ by applying the well known formula of the Bohr model as E1 = 5.39 = 0 - (-13.6) ζ2 . That is ζ = 1.26 > 1 Whereas for the explanation of the second ionization energy E2 = 75.64 eV one must write E2 = E(1s1) - E(1s2) = (-13.6) Z2 - [ (-27.2)Z2 +(16.95)Z - 4.1] Since Z = 3 one gets E2 = (-13.6)32 + (27.2)32 - (16.95)3 + 4.1 = - 122.4 + 244.8 - 50.84 + 4.1 = 75.64 eV Moreover to explain the third ionization energy E3 = 122.4 eV one must write the simple formula based on the Bohr model for Z = 3 as E3 = 0 - (-13.6)Z2 = 0 - (-13.6)32 = 122.4 eV Category:Fundamental physics concepts